A parallel - plate capacitor is being charged.Show that the displacement current across an area in the region between the plates and parallel to it is equal to the conduction current in the connecting wires.

Explanation

Solution: The electric field between the plates is
`E = (Q/(varepsilon_0) A)
where Q is the charge accumulated at the positive plate. `
The flux of this field through the given area is
`Phi_E= (Q/(varepsilon_0) A) xx A = (Q/ varepsilon_0)`
The displacement current is
` (i_d= (varepsilon_0)((d Phi_E)/dt) = (varepsilon_0)(d/dt)(Q/varepsilon_0)) = dQ/dt.`
But dQ/dt is the rate at which the charge is carried to the positive plate through the connecting wire. Thus, `i_d = i_e.`